Chapter 8: Problem 19
What is the electronegativity trend? Where does hydrogen fit into the electronegativity trend for the other elements in the periodic table?
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Chapter 8: Problem 19
What is the electronegativity trend? Where does hydrogen fit into the electronegativity trend for the other elements in the periodic table?
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Which of the following statements is(are) true? Correct the false statements. a. The molecules \(\mathrm{SeS}_{3}, \mathrm{SeS}_{2}, \mathrm{PCl}_{5}, \mathrm{TeCl}_{4}, \mathrm{ICl}_{3}\), and \(\mathrm{XeCl}_{2}\) all exhibit at least one bond angle which is approximately \(120^{\circ}\) b. The bond angle in \(\mathrm{SO}_{2}\) should be similar to the bond angle in \(\mathrm{CS}_{2}\) or \(\mathrm{SCl}_{2}\) c. Of the compounds \(\mathrm{CF}_{4}, \mathrm{KrF}_{4}\), and \(\mathrm{SeF}_{4}\), only \(\mathrm{SeF}_{4}\) exhibits an overall dipole moment (is polar). d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to maximize electron repulsions.
Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part \(\mathrm{c}\), why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2} ?\) \(\mathrm{Mg}_{2} \mathrm{O} ?\)
What is meant by a chemical bond? Why do atoms form bonds with each other? Why do some elements exist as molecules in nature instead of as free atoms?
Look up the energies for the bonds in \(\mathrm{CO}\) and \(\mathrm{N}_{2}\). Although the bond in \(\mathrm{CO}\) is stronger, \(\mathrm{CO}\) is considerably more reactive than \(\mathrm{N}_{2}\). Give a possible explanation.
Use Coulomb's law, $$ V=\frac{Q_{1} Q_{2}}{4 \pi \varepsilon_{0} r}=2.31 \times 10^{-19} \mathrm{~J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right) $$ to calculate the energy of interaction for the following two arrangements of charges, each having a magnitude equal to the electron charge.
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